## Description

This article is from the Puzzles FAQ,
by Chris Cole chris@questrel.questrel.com and Matthew Daly
mwdaly@pobox.com with numerous contributions by others.

# 177 geometry/dissections/largest.circle.p

What is the largest circle that can be assembled from two semicircles cut from

a rectangle with edges a and b?

geometry/dissections/largest.circle.s

There are two methods:

Method M1:

The diameters of the semicircles have to be on the longer sides,

starting at an endpoint of the rectangle. The two semicircles touch

each other in the middle M of the rectangle.

a
D._______________________.C
| |
| |
b | . M |
| |
| |
|_______.___.___________|
A R X B

R should be the center of the semicircle, and because of RA = RM,

it holds that:

r^2 = (a/2 - r)^2 + (b/2)^2

Solving for r gives:

r = min[b,(a^2+b^2)/(4a)], where a >= b.

Method M2:

We'll cut on the line y = c x, where c will turn out to be slightly

less than d, the slope of the diagonal. We describe the semicircle

lying above the line y = c x, having this line as the straight part of

the semi-circle. The center P of the semicircle will be taken on the

line y = d - x, and will be tangent to the left and top of the

rectangle. Clearly the lower down P is on this line, the better. The

naive solution is not optimal because the upper place where the

semicircle meets the diagonal is interior to the rectangle. So we try

to determine c in such a way that this latter point actually lies

slightly down from the top, on the right side of the rectangle. This

involves solving the quartic:

4r^4 - (4a+16b)r^3 + (16b^2+a^2+8ab)r^2 - (6b^3+4ab^2+2ba^2)r + b^4+(ab)^2 = 0,

where r < b, the details of which will be left to the reader.

The other semicircle is the reflection of the first through the origin.

After a few calculations, we find that the value of r given

by M2 is greater than the one given by M1 only if 1 < a/b < 2.472434.

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